TITLE: Without calculator prove that $\log^211+\log^29<\log99$
QUESTION [10 upvotes]: Without calculator prove that $$\log^211+\log^29<\log99$$

The base of logarithm is $10$.
I have a proof for the similar inequality. Prove that:
$$\log^211+\log^29>\log98.$$
Indeed, $$\log^211+\log^29>\frac{1}{2}\left(\log11+\log9\right)^2=\frac{\log^299}{2}=\frac{\log^299^2}{8}=$$
$$=\frac{\log^29801}{8}>\frac{\log^29800}{8}=\frac{(\log98+\log100)^2}{8}>\frac{\log98\log100}{2}=\log98$$

REPLY [5 votes]: Hint: let $f(x)=\log^2(x)-\log(x)$ then $f''(x)=\cfrac{-2 \ln(x) + 2 + \ln(10)}{x^2\, \ln^2(10)}\,$ and $f''(x) \lt 0$ for $x \gt \sqrt{10}\,e$. Given that $\sqrt{10}\,e \lt 9\,$ it follows that $\frac{1}{2}\left(f(9)+f(11)\right) \lt f\left(\frac{1}{2}(9+11)\right) = f(10)=0\,$ by concavity of $f(x)\,$, and the latter is equivalent to the proposed inequality.